Given: The wavelength of a proton or a neutron is 10 −15 m and the rest mass energy of electron is 0.511 MeV.
de Broglie wavelength is given as,
λ= h p
Where, p is the momentum and h is the Planck’s constant.
By substituting the given values in the above expression, we get
10 −15 = 6.6× 10 −34 p p=6.6× 10 −19 kgms -1
The relativistic kinetic energy is given as,
E 2 = p 2 c 2 + m e 2 c 4 = ( pc ) 2 + ( m e c 2 ) 2
Where, E is the kinetic energy, c is the speed of light and m e is the rest mass of an electron.
By substituting the values in the above equation, we get
E 2 = ( 6.6× 10 −34 ×3× 10 8 ) 2 + ( 0.511× 10 6 ×1.6× 10 −19 ) 2 =392.04× 10 −22 E= 392.04× 10 −22 =1.98× 10 −10 J
Further simplifying the above expression, we get
E=1.98× 10 −10 J× 1 eV 1 .6×10 -19 J =1.24× 10 9 eV
Thus, the electron energy emitted from the accelerator at Stanford, USA will be of the order 1.24× 10 9 eV.